Hollinger Corp. 
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LC 6301 
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fIrit¥ersitY and SchsQl Extensien. 

GEOMETRY— Course B. 

A. W. PHILLIPS, 

1889. 

Yale University. 



u 






Copyright. 

1889, 

By a. W. Phillips. 



Press of J. J. Little & Co. 
Aster Place, New York. 



Geometry.-Course B. 
solid geometry. 

OBJECTS OF THE COURSE. 

First— 1:0 enable the student to make the constructions and 
solve the problems of Mensuration, which depend upon the princi- 
ples of elementary Solid Geometry. 

Second— 1:0 enable the student to demonstrate the principles 
upon which these constructions and solutions depend. 

METHODS OF STUDY. 

The student is advised to study the definitions and demonstra- 
tions found in the text-book of solid geometry which he has used in 
plane geometry. He should be able to draw the figures required for 
the problems of the syllabus in accordance with the methods sug- 
gested. He should then solve the problems in the syllabus, and 
write out an outline of the demonstrations of the propositions in- 
volved. 

MODELS. 

It is recommended that in the instruction of a class, or where 
several persons study together, that, especially in the early part of 
the work, models of the figures be made of wood, pasteboard, metal. 



UNIVERSITY EXTENSION. 



wire, strings, etc., so that a clear conception may be formed of the 
figure in space. 

DRAWINGS. 

Each student should then learn how to make the drawings 
mathematically from measurements taken from the models. The 
simplest means of drawing the figures is in what is called " cabinet 
projection," several examples of which are given in the following 
pages. 

In the particular system which is most fully explained here, the 
drawings may be regarded as the shadows of the skeleton outlines of 
geometrical models, as they would be cast by the sun on a vertical 
plane. The position of the sun is assumed to be such that all lines 
in the model perpendicular to the vertical plane are represented in 
the drawing by lines of one half their length, and inclined at an 
angle of sixty degrees to horizontal lines in the drawing. 

In this system : 
1°. All vertical lines in the model are vertical lines in the drawing. 
2°. All horizontal lines in the model parallel to the vertical plane are 

horizontal lines in the drawing. 
3°. All lines in the model pei^pendicular to the vertical plane are 

inclined at an angle of 6o° to horizontal lines in the drawing. 
4°. All lines parallel to the vertical plane in the model have the 

same length in the drawing. 
5°. All lines perpendicular to the vertical plane in the model are 

one half as long in the drawing. 
(In making any drawing, proportional dividers may be used 



GEOMETRY.— COURSE B. 



B 


/ 




/ 










/ 




K 










Fig. I. 



to lay off the lines perpendicular to the 
vertical plane.) 

Figure i illustrates the method of 
drawing a cube whose edges are either 
parallel or perpendicular to the vertical 
plane on which the shadow would be cast. 
From any point O, draw the vertical line 
O B and the horizontal line O A, each equal 
in length to the edge of the cube. Make 
the angle A O C equal to 60°, and O C 

equal to half the length of the cube. Complete the figure by 
drawing the remaining edges of the cube parallel respectively to 
those already drawn. 

Figs. 2 and 3 illlustrate the method of drawing a pyramid 
whose base is in the horizontal plane, but the sides of the base are 
neither parallel nor perpendicular 
to the vertical plane. 

The quadrilateral in Fig. 2 rep- 
resents the actual size and position 
of the base. O/ is a line drawn 
in the plane of this base parallel to 
the vertical plane, and the several 
perpendiculars drawn to the line 
0/ from the corners of the base, 
and meeting it in ^ ^^ d f, show what 
kind of lines must be drawn in any 
horizontal figure in order to lay it off in the projection here de- 




FiG. 2. 



UNIVERSITY EXTENSION. 



scribed. A is the position of the foot of a perpendicular drawn 
from the vertex of the pyramid to the base. 



£ 



M 


H 

i 

■"7" 


/ / / 


«- 5 


Fig. 


d f 

3- 



Fig. 3 is the pro- 
jection of the pyramid. 
O/is a horizontal line 
with points a b c d f 
marked upon it in the 
same scale as the cor- 
responding line of Fig. 
2. The perpendiculars 
(S ' / / / f- / / in Fig. 2, drawn from 

abed, etc., to the cor- 
ners of the base, are 
drawn in Fig. 3 at an angle of 60° to the line O /, and on a scale 
of one half. The extremities of these lines are joined to make the 
base of the figure. At A the vertical line A B is drawn of the 

same height as the pyramid, 
and the vertex B is joined to 
each one of the corners of the 
base. The large parallelogram 
represents the projection of a 
rectangular board on which the 
pyramid stands. 

Figs. 4 and 5 illustrate how 
the horizontal circular base of a 
figure may be drawn in this pro- 
jection. Let the circle in Fig. 




GEOMETRY.— COURSE B. 



this 



4 be the base of a figure, as it stands, with respect to a horizontal 

line of reference, as O g, parallel to the vertical plane of projection. 

Draw a horizontal diameter 

to the circle. Divide 

diameter into a 

equal parts, and throu< 

several points of division A X c ^ 

draw chords to the circle, pj^ ^ 

which shall be perpendicular 

to O g, and meet that line in ^ <^ ^, etc. Fig. 5 is the projection of 

Fig. 4. 

In Fig. 5, O^ is a horizontal line of the same length and 
divided in the same way as the corresponding line of Fig. 4. At 



Z r / / /J 
,' ^^ — » 



^ f S 



each of the points of division a d c, etc., a line is drawn, making an 
angle of 60° with O g. The segments on each of these lines between 
O g and the curve and its 
horizontal diameter are 
half the corresponding 
segments in Fig. 4. After 
laying off these segments 
the points are joined. 

Fig. 6 illustrates how 
any simple model may be 
drawn in this projection. 
The large parallelogram 
is the projection of a rect- 
angular horizontal board. 




8 



UNIVERSITY EXTENSION. 



AB is a rod perpendicular to ttiis board through a given point; 
D C, a Hne of the board parallel to one of its edges. O E is per- 

C/ 





pendicular to D C, and A C B, A E B, and A D B are strings join- 
ing A and B, and passing through the board at D, E, and C. The 
lines A B and D C, and the sides of the parallelogram parallel to 
D C, are the same length as in the model. O E is inclined to D C 

60°, and is drawn on the scale one half. 
The other lines, being neither parallel nor 
perpendicular to the vertical plane of pro- 
jection, are not drawn to a scale, but sim- 
ply connect points established by lines 
parallel or perpendicular to the vertical 
plane of projection. 

Figs. 7, 8, 9 represent the outlines of 
a cylinder, cone, and sphere in this pro- 
jection. The bases of the cylinder and 




Fig. 9. 



GEOMETRY. — COURSE B. 



cone and the horizontal section of the sphere may be drawn on the 
same principle as the projection of the circle in Fig. 5. 



\ 




\ 


p 








\ 



B 



A 



y 

"A 



Fig. 10. 



Fig. II. 



Figs. 10 and 11 are constructed on the principle that all lines 
either parallel or perpendicular to the vertical plane of projection 
are of the same length in the drawings as in the models. The lines 
CO are inclined 45° or 135" to the line O A. It maybe said in 
general that the lines perpendicular to the vertical plane may be 
inclined at any angle whatever to the horizontal lines in the draw- 
ing, and may be made on any scale whatever. One very convenient 
arrangement is to make the angle 45° and draw the lines so inclined 
on a scale of one-half. 

Fig. 12 is an example of what is called isometric projection. In 
this system it is assumed that the projection of a cube is made by a 
shadow cast on a plane perpendicular to the direction of the rays 
of light, and that the rays are in the direction of one of the longest 
diagonals of the cube. The lines, therefore, to be measured and 



lO 



UNIVERSITY EXTENSION. 




laid off for any figure are the 
vertical lines, the lines perpendic- 
ular to some fixed vertical plane, 
and the lines parallel to horizon- 
tal lines in that vertical plane. 
The scale of all these lines is 
the same, and they are laid off 
respectively parallel to O B, O A, 
and O C in the figure. The 
angles A O B and B O C are each 
60°. 



BOOK I. 



LINES AND PLANES IN SPACE 



1. Can more than one plane be passed through 

(a) A line ? (c) Two intersecting lines ? 

(d) A line and a point? {d) Tv/o parallel lines? 
What is the intersection of two planes ? 

2. Why will a three-legged stool always stand firm on a plane floor ? 

3. Can a person determine by a straight-edge whether a surface is 

a plane or not ? In what ways ? 

4. How would a person determine whether a post in a room was 

perpendicular to the floor ? Could he determine it by two 
carpenter's squares ? 



GEOMETRY.— COURSE B. II 

5. How would one find where a perpendicular from a point in the 

ceiling would strike the floor by the use of a ten-foot pole, the 
height of the ceiling being less than 10 feet ? 

6. Suppose a straight line A B is marked on the floor : At its two 

extremities, A and B, a string is fastened, the string being 
longer than A B. How could a plane perpendicular to A B 
at its middle point be determined ? 

7. Suppose a rectangular door is swung on a vertical post : How 

can one tell whether the floor is level or not ? 

8. When the door in the above example swings, is it parallel in all 

positions to every vertical line in the room ? 

9. A table whose top is level stands on a level floor. Why is the 

edge of a ruler lying anywhere on the top of the table parallel 
to the floor ? 

10. Suppose any straight line drawn obliquely across a room. From 

a point in the hinge of a desk-lid in the room how would one 
draw a line parallel to the first line ? How would the desk-lid 
be made parallel to the first line ? 

11. The floor and ceiling of a room are parallel planes. Why is a 

post perpendicular to one perpendicular to the other ? 

12. The floor and ceiling of a room are both perpendicular to a post. 

Why are they parallel to each other ? 

13. Why are the floor and ceiling of a room (parallel planes) every- 

where equidistant ? 

14. Two parallel lines intersect two parallel planes. Why are they 

equal ? 

15. A line A B cuts three parallel planes in A, F, and B, and a line 



12 UNIVERSITY EXTENSION. 

C D cuts the same planes in C, G, and D. If A F = 3, F B == 
4, and C D = 6, compute C G. 

16. Prove that two angles not in the same plane, whose sides are re- 

spectively parallel and lie in the same direction, are equal, and 
that their planes are parallel. 

17. How may the diedral angle between two planes be measured ? 

For example, the angle between any two faces of a pyramid : 
the angle between two faces of a crystal. 

18. How may a plane surface be set up so as to make an angle of 

50° with the floors ? How may two plane surfaces be set at 
right angles to each other ? How made parallel? 

19. How shall the angle which a line makes with a plane be 

measured? How shall a line be drawn which shall be a 
perpendicular common to any two non-parallel lines in space ? 

20. When are two triedral angles equal to each other ? 

21. Prove that the sum of the face angles about the vertex of any 

pyramid is less than four right angles. 



GEOMETRY. — COURSE B. 1 3 

BOOK II. 

SOLIDS BOUNDED BY PLANES. 

THE COVERINGS. 

22. Find the amount of paper needed to cover the walls and ceiling 

of a room 18 feet long, 16 feet wide, and to feet high. 

23. A building 16 feet square has a roof in the form of a square 

pyramid. The altitude of this pyramid is 10 feet. Required 
the surface of the roof. 

24. The side of the base of a square pyramid measures 10 feet, and 

the distance from each corner of the base to the vertex is 15 
feet. Required the surface of the roof. 

25. The base of a pyramid is a regular hexagon inscribed in a circle 

of 5 feet radius. The slant height of this pyramid is 8 feet. 
Find the total surface of the pyramid including the base. 

26. A rectangular building 12 feet wide by 20 feet long has a roof in 

the form of a wedge. The altitude of this roof is 10 feet. 
The ridge-pole of the roof is 8 feet in length. Required the 
surface of the roof. 

27. A bin is in the form of the frustum of a square pyramid in- 

verted. The bottom is 3 feet square, the top 9 feet square, 
the depth 4 feet. What is the number of square feet of cov- 
ering required for the bottom and sides ? 

28. An oblique pyramid has a square base the side of which is 12 

inches. The altitude of the pyramid is 8 feet, and the vertex 



UNIVERSITY EXTENSION. 



is in the same vertical line with one corner of the base. Re- 
quired the surface of the sides of the pyramid. 

29. Cut out from pasteboard the covering of the following regular 

polyedrons : 

(a) When an edge is equal to unity, 

(l>) When the radius of the circumscribed sphere is unity, 
for as many as can be found by geometry. 
Tetraedron. ■ Octaedron. Icosaedron. 

Cube. Dodecaedron. 

30. How would the diedral angle of the edge be measured ? 

THE VOLUMES. 

31. Find the contents of a right parallelopiped. 

32. When are two solids similar ? 

^^. Find the contents of a right triangular prism, the sides of the 
base being 3, 4, and 5 inches respectively, and the altitude 
being 8 inches. 

34. Prisms having equivalent bases are to each other as their alti- 

tudes. 

35. Two prisms which have the same height are to each other as 

their bases. 

36. Find the contents of an oblique prism whose base is a rectangle, 

whose sides are 3 and 6 inches, and whose altitude is 3 inches 
— the length of the oblique edge is 4 inches. Show, by taking 
a package of paper cut this size, that it may be deformed 
from a right parallelopiped into an oblique one in two direc- 
tions. The altitude and base will of course remain the same. 



GEOMETRY. — COURSE B. 



15 



37. Prove that two pyramids having equal bases and the same alti- 
tude have the same volume, by the " method of limits." 
Illustrate this proposition by packages of paper. 

^8. Show the same also when the bases are equivalent^ but of differ- 
ent shape. 

39. Show that when two pyramids have the same altitude they are 

to each other as their bases. 

40. Show that, when a pyramid is cut off by a plane parallel to its 

base, the part cut off is similar to the whole. 

41. Two similar prisms or pyramids are to each other as the cubes 

of their homologous sides. 

42. A cube can be cut into six equal square pyramids having their 

bases the faces of the cube and their vertices in the center. 
Show by this that the volume of the pyramid is equal to the 
base multiplied by one-third the altitude. 

43. Show that the volume of any pyramid is equal to the product of 

the base by one-third the altitude. 

44. Show that the volume of the frustum of a regular square pyra- 

mid is equal to 1 [B^ + b'^ + Bb) h, when B and b and h are 
respectively the sides of the lower base, the side of the upper 
base, and the altitude. 

[This may be worked by algebra, using the principle of similar pyramids. 
Find first the altitude of the complete pyramid. Then subtract the volume of 
what was cut off from the volume of the complete pyramid, and reduce the expres- 
sion for the remainder to the above form,] 

45. Make of wood a triangular prism of any convenient size, and 

then cut it into three triangular pyramids, Show that these 
pyramids are equivalent to each other. 



l6 UNIVERSITY EXTENSION. 

Note. — The student can find numerous examples in the arithmetics as exer- 
cises in the rules of practical mensuration. Especial attention should be given to 
examples involving the flooring, plastering, and papering of rooms, the amount of 
brick or stone needed in constructing the walls, the lumber used in the roof, etc.; 
also, the surface and capacity of cisterns and bins, and problems involving what is 
called "board measui-e." 



BOOK III. 

THE CONE AND CYLINDER. 

46. The convex covering of the right cylinder is equal to a rectangle 

whose length is the length of the cylinder, and whose width is 
the circumference of the cylinder. 

47. The surfaces of two cylinders are to each other as 4 to 16 ; the 

radius of the base of the first is 6 inches. Required, the con- 
vex surfaces of both cylinders. 

48. The convex covering of the right cone is a sector of a circle 

whose radius is the slant height of the cone, and the angle at 
the center is less than a circumference. 

49. Compute the surface of a right cone, the radius of the base 

being i inch, and the slant height 4 inches, and cut out the 
covering from paper ; also, compute the angle of the sector 
of the circle and cut out the covering, supposing the altitude 
of the cone to be 2 inches, and the radius of the base ij 
inches. 

50. Calculate the surface of the frustum of a right cone, the radius 

of whose upper base is i inch, of the lower base 2^ inches, 
and the slant height 2^ inches. 



GEOMETRY. — COURSE B. 1 7 

51. Prove that the volume of a cylinder is equal to the area of the 

base multiplied by the altitude. 

52. Prove that the volume of a cone is equal to one-third the product 

of the base by the altitude. 

53. Prove that the volume of the frustum of a cone is equal to 

^ {A -}- a + \/Aa) h where A and a are respectively the areas 
of the lower and upper bases, and h the altitude of the frustum^. 
[The same method may be employed as in No. 44.] 
Note. — The student should apply these rules to the problems of mensuration. 

54. The area of the base of a right cone is 3 square feet, and its 

height 30 inches. Find the height of a cone the solid con- 
tents of which is four times as great, but the diameter of 
whose base is only one-third of the given one. 



BOOK IV. 

THE SPHERE. 



[In studying the sphere it is of advantage to use a slated globe, on which the 
figures which relate to the surface may be constructed. The globe should be fitted 
to a cylindrical cup such that the globe can be turned easily in the cup, and of a 
depth equal to the radius of the sphere. The rim of the cup may then be used as 
a ruler for drawing the arcs of great circles on the sphere.] 

PROBLEMS TO BE CONSTRUCTED ON THE GLOBE. 
55. {a) Bisect a given arc. [b) Erect a perpendicular to a given 
great circle from a point on the great circle, and also from a 
point outside. 



UNIVERSITY EXTENSION. 



56. (a) Construct an angle equal to a given angle, (l?) Construct a 

triangle given the three sides, (c) Construct a triangle given 
two sides and the included angle, (d) Construct for all pos- 
sible cases a triangle given two sides and an angle opposite 
one of them. 

57. On the sphere, determine by the dividers the length of the diam- 

eter. 

58. Explain the methods of measuring the angle of any spherical 

triangle. 

59. What are the methods of proving two spherical triangles equal ? 

60. Derive the rule for finding the surface of a sphere. 

61. Compute the surface of a sphere whose radius is 6 inches. What 

is the radius of a sphere whose surface is one half that of the 
above ? 

62. Find the surface of the earth, supposing its radius to be 4,000 

miles. Derive the rule for finding the convex surface of a 

zone of the earth given the altitude. 

6^. Prove that the area of a spherical triangle is equal to its spherical 

excess, multiplied by one-eighth of the surface of the sphere. 

[We compute the area of a spherical triangle by adding the angles together, 
subtracting from their sum 180°, and dividing the remainder by 90°. Then 
multiply the resulting fraction hjjTt R'-^ for the area.] 

64. Compute the volume of the earth, and also that of a zone. 



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